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How do I find the intersections for the equations, $y = 4-x^2$ and $y = 2^x + 1$ without drawing or using a graphing calculator?

I got this far:

$x^2 + 2^x - 3 = 0$

Tried using some calculating programs but they could not compute the answer, and I could only find the x and y values by using a graph calculator. How can I find the x and y values without one?

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    $\begingroup$ You can use an approximation method like the newton-raphson method: en.wikipedia.org/wiki/Newton%27s_method $\endgroup$ – callculus Sep 27 '15 at 15:19
  • $\begingroup$ On the other hand you can take $x=1$, because $1^2+2^1-3=0$. It is more or less obvious. $\endgroup$ – callculus Sep 27 '15 at 15:23
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Continuing from @calculus' comments, consider the function $$f(x)=x^2 + 2^x - 3$$ As noticed, $x=1$ is one solution.

On the other hand, $f(0)=-2$ and $f''(x)=2^x \log ^2(2)+2 \gt 0$; so there is another root. Using inspection $f(-1)=-\frac{3}{2}$, $f(-2)=\frac{5}{4}$ which shows a solution between $-2$ and $-1$.

Now, start Newton method, say with $x_0=-\frac{3}{2}$. This will produce as successive iterates $-1.64390$, $-1.63659$, $-1.63658$ which is the solution for six significant figures.

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You should be able to look at the function and tell that $x=1$ is a solution. As for the other point of intersection it would be very hard by hand, but if you work with a scientific calculator, as already stated you may use Newtons method.

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

$x_1$ is a guess to what might be the solution. I would guess a negative number. As we get $x_2, x_3,x_4.....$ we get closer and closer to the answer (in terms of only $x$) you're looking for.

Note that $f'(x_n)$ is the derivative of a function, for this function:

$$f'(x_n)=2x_n+ln2(2^{x_n})$$

And so this is the formula we use to get better and better guesses:

$$x_{n+1}=x_n-\frac{(x_n)^2+2^{x_n}-3}{2x_n+ln2(2^{x_n}) }$$

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GeoGebra will solve it (in the CAS View) quite happily

NSolve[x^2 + 2^x - 3 = 0]

{x = -1.636576038445, x = 1}

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